How do I solve this three variable equation
2 answers:
A) 2.x - 6.y + z = 30 b) -x + 3.y +2.z = - 20 c) -6.x -5.y +6.z = 7 1st, let's take a) and b) alone and let's eliminate x a) 2.x - 6.y + z = 30 b) -x + 3.y +2.z = - 20 2nd) multiply b) with 2 and once done add a) with the result of b): b') times 2 → -2.x+6.y+4.z = - 40. Now add it to a): a) 2.x - 6.y + z = 30 b') -2.x+6.y+4.z = - 40 + ----------------- ----- 0.x + 0.y + 5.z = -10 → z = -2 Now replace z with its value (-2) in : b) -x + 3.y +2.z = - 20 → - x + 3.y -4 = -20 → - x +3.y = -16 c) -6.x -5.y +6.z = 7 → -6.x - 5.y -12 = 7 → - 6.x -5.y = 19 Consider the result of: - x +3.y = -16 and - 6.x -5.y = 19 Now let's eliminate x from both equations. To that end multiply the 1st one by (-6) 6x - 18y = 96 and add it to - 6.x -5.y = 19 6x - 18y = 96 -6.x -5.y = 19 +--------- ---- 0.x - 23.y 115 → 23y = -115 and y = -115/23 = - 5→→ y = -5 Now that you have the value of z= - 2 and y = -5, plugin in any of the equation and you will find x = 1. So the final result is: x = 1 ; y = - 5 ; and z = - 2
2x - 6y + z = 30 (1st) -x + 3y + 2z = -20 (2nd) -6x - 5y + 6z = 7 (3rd) from (2nd) you have: -x + 3y + 2z = -20 x = 3y + 2z + 20 substitute x = 3y + 2z + 20 into (1st) 2x - 6y +z = 30 2x - 6y + z = 30 2(3y + 2z + 20) - 6y +z = 30 6y + 4z + 40 - 6y + z = 30 5z + 40 = 30 5z = -10 z = -2 plug in z = - 2 into (2nd) and (3rd) -x + 3y + 2(-2) = -20 (2nd) -x + 3y = -16 so x = 3y + 16 -6x - 5y + 6z = 7 (3rd) -6x -5y +6(-2) = 7 -6x - 5y = 19 substitute x = 3y + 16 into -6x - 5y = 19 -6x - 5y = 19 -6(3y + 16) - 5y = 19 -18y - 96 - 5y = 19 -23y = 115 y = -5 x = 3y + 16 x = 3(-5) + 16 x = -15 + 16 x = 1 answer x = 1, y = -5 and z = -2
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